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URL: https://oeis.org/A120972

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A120972
G.f. A(x) satisfies A(x/A(x)^3) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^3).
9
1, 1, 3, 21, 217, 2814, 42510, 718647, 13270944, 263532276, 5567092665, 124143735663, 2905528740060, 71058906460091, 1809695198254281, 47861102278428198, 1311488806252697283, 37164457324943708739, 1087356593493807164289, 32801308084353988297404
OFFSET
0,3
COMMENTS
More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.
FORMULA
G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^3.
a(n) = [x^(n-1)] A(x)^(3*n)/n for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in A(x)^(3*n)/n for n>=1 (see comment).
Let B(x) be the g.f. of A120973, then B(x) and g.f. A(x) are related by:
(a) B(x) = A(A(x)-1),
(b) B(x) = A(x*B(x)^3),
(c) A(x) = B(x/A(x)^3),
(d) A(x) = 1 + x*B(x)^3,
(e) B(x) = 1 + x*B(x)^3*B(A(x)-1)^3,
(f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^3).
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120973.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(3*n+k,j)/(3*n+k) * b(n-j,3*j).
a(n) = b(n-1,3) for n > 0. (End)
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
Related expansions.
A(x)^3 = 1 + 3*x + 12*x^2 + 82*x^3 + 813*x^4 + 10212*x^5 + 150699*x^6 +...
A(A(x)-1) = 1 + x + 6*x^2 + 60*x^3 + 776*x^4 + 11802*x^5 + 201465*x^6 +...
A(A(x)-1)^3 = 1 + 3*x + 21*x^2 + 217*x^3 + 2814*x^4 + 42510*x^5 +...
x/A(x)^3 = x - 3*x^2 - 3*x^3 - 37*x^4 - 420*x^5 - 5823*x^6 -...
Series_Reversion(x/A(x)^3) = x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(3*n)/(3*n),
form a table of coefficients in A(x)^(3*n) as follows:
A^3: [(1), 3, 12, 82, 813, 10212, 150699, 2503233, ...];
A^6: [ 1, (6), 33, 236, 2262, 27270, 388906, 6289080, ...];
A^9: [ 1, 9, (63), 489, 4671, 54684, 756012, 11904813, ...];
A^12: [ 1, 12, 102, (868), 8445, 97260, 1310040, 20112516, ...];
A^15: [ 1, 15, 150, 1400, (14070), 161343, 2130505, 31961175, ...];
A^18: [ 1, 18, 207, 2112, 22113, (255060), 3324003, 48876264, ...];
A^21: [ 1, 21, 273, 3031, 33222, 388563, (5030529), 72769014, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[3/3*(1), 3/6*(6), 3/9*(63), 3/12*(868), 3/15*(14070), 3/18*(255060), ...].
MATHEMATICA
terms = 18; A[_] = 1; Do[A[x_] = 1 + x*A[A[x] - 1]^3 + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 15 2018 *)
PROG
(PARI) {a(n)=local(A=[1, 1]); for(i=2, n, A=concat(A, 0); A[ #A]=-Vec(subst(Ser(A), x, x/Ser(A)^3))[ #A]); A[n+1]}
(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=1+x*subst(A^3, x, A-1+x*O(x^n))); polcoeff(A, n)}
(PARI) /* This sequence is generated when k=3, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=3, m=0)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}
for(n=0, 25, print1(a(n), ", "))
(PARI) b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(3*n+k, j)/(3*n+k)*b(n-j, 3*j)));
a(n) = if(n==0, 1, b(n-1, 3)); \\ Seiichi Manyama, Jun 04 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 20 2006
STATUS
approved